construct a APOR in which QR=6.5cm,Angle Q=60°
and PR - PQ=115cm and iustify
it
Answers
Given, in ∆PQR,
Given, in ∆PQR,QR = 6.5 cm, ZQ = 60° and
Given, in ∆PQR,QR = 6.5 cm, ZQ = 60° and PR - PQ =1.5 cm
Given, in ∆PQR,QR = 6.5 cm, ZQ = 60° and PR - PQ =1.5 cmSteps of construction
Given, in ∆PQR,QR = 6.5 cm, ZQ = 60° and PR - PQ =1.5 cmSteps of construction(i)Draw the base, QR = 6.5 cm
i.e. the side containing the base angle 0 is less than third side, so it is the case II.
(ii)Cut line segment OS equal to PR-PQ,
i.e.OS =1.5 cm from the ray OX extended on opposite side of base QR.
(iii)Join SR and draw its perpendicular bisector ray AB, which intersects SR at M (say).
(iv)Let P be the intersection point of SX and perpendicular bisector AB. Then, join PR.
(iv)Let P be the intersection point of SX and perpendicular bisector AB. Then, join PR.image
Thus, ∆PQR is the required triangle.
Justification
JustificationBase QR and ∠Q are drawn as given.
Since, AB is the perpendicular bisector of SR and P lies on it.
Since, AB is the perpendicular bisector of SR and P lies on it.∴ PS = PR
Since, AB is the perpendicular bisector of SR and P lies on it.∴ PS = PRNow, QS = PS - PQ
Since, AB is the perpendicular bisector of SR and P lies on it.∴ PS = PRNow, QS = PS - PQ=> QS= PR-PQ
Thus, construction justified.
Answer:
Given, in ∆PQR,
QR = 6.5 cm, ZQ = 60° and
PR - PQ =1.5 cm
Steps of construction
(i)Draw the base, QR = 6.5 cm
image
image
i.e. the side containing the base angle 0 is less than third side, so it is the case II.
(ii)Cut line segment OS equal to PR-PQ, i.e.
OS =1.5 cm from the ray OX extended on opposite side of base QR.
(iii)Join SR and draw its perpendicular bisector ray AB, which intersects SR at M (say).
(iv)Let P be the intersection point of SX and perpendicular bisector AB. Then, join PR.
image
Thus, ∆PQR is the required triangle.
Justification
Base QR and ∠Q are drawn as given.
Since, AB is the perpendicular bisector of SR and P lies on it.
∴ PS = PR
Now, QS = PS - PQ
=> QS= PR-PQ
Thus, construction justified.